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We're currently learning about applying the TISE to one-electron (hydrogen) atoms in my intro to QM class. While reading about it in the textbook, I was a bit confused about radial probability density and distribution functions; mainly as my textbook uses them somewhat interchangeable (which I don't necessarily think is advised). Thus, I was wondering if I could receive some answers regarding to the following.

  1. What is the difference between a radial distribution function and a radial density function? I checked a similar question What is a good definition for Radial Probability Density?. but am still confused between the actual difference between them. For example, when would I use one vs. the other? Why?

  2. Moreover, relating to the actual integrals used in solving for the radial distribution function, why does the first (radial) integral just disappear? Again, I looked at this same thing here Calculating the Radial Distribution Function and the answer mentioned something regarding Riemann sums which, again, I didn't quite understand.

I do apologize if some of these questions/clarification are rather trivial, however, any assistance would be greatly appreciated.

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    $\begingroup$ The question you link to says: "The radial distribution function is just $r^2|R_{nl}(r)|^2$, so it is the radial probability density multiplied by $r^2$." Does this answer your first question? $\endgroup$
    – Andrew
    Apr 4 at 4:53
  • $\begingroup$ @Andrew Kind of? What I'm essentially trying to ask is, "What is the difference between ๐‘Ÿ^2|๐‘…๐‘›๐‘™(๐‘Ÿ)|^2 and 4(pi) ๐‘Ÿ2|๐‘…๐‘›๐‘™(๐‘Ÿ)|2?" and why are they used somewhat interchangeably even when I'm quite sure that they do not refer to the same thing? For example, here: chem.libretexts.org/Bookshelves/… they define the quantities as two separate things, yet I'm unclear as to what's the difference between them. $\endgroup$
    – m_1265
    Apr 5 at 18:59

2 Answers 2

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Setup

Let's focus on the ground state for concreteness. The wave function is \begin{equation} \psi_0(\vec{r}) = \frac{1}{\sqrt{\pi a_0^3} } e^{-|\vec{r}|/a_0} \end{equation} where $a_0$ is the Bohr radius.

Probability density function

Given the wavefunction, we can compute a three dimensional probability density function (pdf) $P(\vec{r})$ \begin{equation} P_{3D}(\vec{r}) = |\psi_0(\vec{r})|^2 = \frac{1}{\pi a_0^3} e^{-2 |\vec{r}| / a_0} \end{equation} Note that the dimension of $P_{3D}$ is $({\rm length})^{-3}$.

Now what does this pdf tell us? Given a small volume of space, whose center is separated from the origin by vector $\vec{r}$, the probability that the electron is contained in that volume is $P_{3D}(\vec{r}) dV$, where $dV$ is the volume of that small volume element.

We can check that if we add up the probability associated with every volume element in space, we get $1$ (since we started from a properly normalized wavefunction).

\begin{eqnarray} \int dV P_{3D}(\vec{r}) &=& \int_0^\infty dr r^2 \int_0^\pi d\theta \sin \theta\int_0^{2\pi} d\phi \frac{1}{\pi a_0^3} e^{-2r/a_0} \\ &=& 4\pi \times \frac{1}{\pi a_0^3} \int_0^\infty dr r^2 e^{-2r/a_0} \\ &=& 4\pi \times \frac{1}{\pi a_0^3} \times \frac{a_0^3}{4} \\ &=& 1 \end{eqnarray}

Radial probability density function

Now, sometimes, instead of wanting to know the probability of finding the electron in a small volume, we want to know what is the probability of finding the electron at some distance from the center. This is equivalent to asking about the probability of the electron being in a spherical shell of radius $r$.

You can derive this in various ways, but the upshot is that you derive a one-dimensional radial probability density $P_{1D}(r)$, which is related to the three-dimensional probability density $P_{3D}(\vec{r})$, via \begin{equation} P_{1D}(r) = 4\pi r^2 P_{3D}(\vec{r}) = \frac{4}{a_0^3} r^2 e^{-2 r/a_0} \end{equation} Note that the dimension of $P_{1D}$ is $({\rm length})^{-1}$ (which is different from the dimension of $P_{3D}$).

The interpretation of this function is that the probability to find an electron between distances $r$ and $r+dr$ from the origin is given by $P_{1D}(r)dr$.

We can check that the probability of finding the electron at some distance is $1$ \begin{eqnarray} \int_0^\infty dr P_{1D}(r) &=& \frac{4}{a_0^3}\int_0^\infty dr r^2 e^{-2 r/a_0} \\ &=& \frac{4}{a_0^3} \times \frac{a_0^3}{4} \\ &=& 1 \end{eqnarray}

Let's also return to $P_{\rm 3D}(\vec{r})$ and see another way to interpret it. We said $P_{\rm 3D}(\vec{r}) dV$ was the probability to find the electron in an infinitesimal volume element with volume $dV$. Since $dV=r^2 dr d\Omega$, where $d\Omega=\sin\theta d\theta d\phi$ is an infinitesimal angular area element on the surface of a sphere, you could also express this probability as $P_{\rm 3D}(\vec{r}) r^2 dr d\Omega = P_{\rm 1D}(r) dr \frac{d\Omega}{4\pi}$. This is the same as the probability of finding the electron in a spherical shell of radius $r$, up to the normalization factor of $d\Omega/4\pi$ that converts the entire surface area of the shell to the infinitesimal cross section associated with the volume element with volume $dV$.

To confuse things futher, sometimes people may be lazy about normalizations, and, say, forget to put in a factor of $4\pi$ but still have in mind a radial pdf. Usually this is ok in practice because when one looks at plots of the probability density, one is comparing relative values and not absolute values. Hopefully, a good and careful resource will not do this.

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For the first question you have, here is my best suggestion.

I'd need to build from a probability, which is the possible over something else (let say, the possible number of electrons in a given space). In the hydrogen atom, it is usually the solution to Schrodinger'S equation.

Then there is the probability density, which is the square of the probability, which is the probability to find an electron in a given state in a given space.

And finally, the distribution. This last one would be the integral over the space of the density, giving you the expected distribution of the electrons in a given state in a given space.

We do use the density instead of the distribution sometimes as knowing the distribution doesn't give us much more information and/or out of laziness or hurry.

Hope it helps, cheers.

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